Thermodynamics Test - 3

A single fixed point chosen as the basis of temperature scale is the Kelvin scale. The state in which ice, liquid water, and water vapor coexist in equilibrium, a state known as the triple point of water, provides the standard reference temperature. The temperature of the triple point of water, which can be very accurately and reproducibly measured, was assigned the value of 273.16 K, corresponding to 0.01°C, in order to maintain the magnitude of a unit temperature.

W = ∫ P dV = PΔV
ΔV = W/P = 54/600 = 0.09 m³
V2 = V1 + ΔV = 0.01 + 0.09 = 0.1 m³.

Concept:

The power required by the fan will be equal to the rate of change of kinetic energy of the air which is caused by the fan.

The rate of change of kinetic energy of the air

\(dE = \frac{1}{2} \times \dot m \times {V^2}\)

where ṁ is the mass flow rate and V is the velocity of the air.

ṁ = ρ × Q

Calculation:

Given,ρ = 1.2 kg/m3, Q = 5 m3/s, V = 10 m/s

Assuming no heat loss in the fan, the power supplied to the fan 

\(dE = \frac{1}{2} \times \dot m \times {V^2}\)

\(dE = \frac{1}{2} \times 1.2 \times5 \times {10^2}=300\;W\)

Concept:

Work done in a reversible isothermal process is given by 

\(W = C\ln \frac{{{P_1}}}{{{P_2}}} = C\ln \frac{{{V_2}}}{{{V_1}}}\)

Where C = P₁V₁ = P₂V₂

P is the pressure and V is the volume of the gas. 

Calculation:

Given, P1 = 1 bar, V1 =  5 m3 and P₂ = 5 bar 

C = P1V1 = 1 bar × 5 m3 = 100 kPa × 5 m3 = 500 kJ

\(W = 500\ln \left( {\frac{1}{5}} \right) = -804.718 \;kJ\)

-ve sign shows that work is done on the system.

Concept:

For the reversible heat engines, efficiency is the function of temperature limits only. If two heat engines have the same heat input and output then their efficiencies must be the same

\(\eta = 1 - \frac{{{T_H}}}{{{T_L}}}\)

where T_H is the higher temperature limit and T_L is the lower temperature limit

Calculation:

Given, \({\eta _I} = {\eta _{II}}\)

For the first engine, TH = 1000 K and TL = T₂ K_, then, 

\({\eta _I} = 1 - \frac{{{T_2}}}{{1000}}\)

For the second engine, TH = T2 K and TL = 400 K, then, 

\({\eta _{II}} = 1 - \frac{{400}}{{{T_2}}}\)

Then, from

\({\eta _I} = {\eta _{II}}\)

\(1 - \frac{{{T_2}}}{{1000}} = 1 - \frac{{400}}{{{T_2}}}\)

Change of internal energy from A to B along path ACB = 180 - 130 = 50 kJ. It will be same even along path ADB.

∴ Heat flow along ADB = 40 + 50 = 90 kJ.

Concept:

From the conservation of energy, Energy of the mixture = Sum of the energy of the individual streams. 

The energy carried by flowing steam = \(mass\;flow\;rate × \frac{{enthalpy}}{{unit\;mass}}\)

Calculation:

Given, ṁ1 = 36 kg/min, ṁ2 = 14 kg/min, h₁ = 36 kJ/(kg da), h₂ = 50 kJ/(kg da)

Energy of the stream 1 = ṁ1 × h1 = 36 × 36 = 1296 kJ/(min da)

Energy of the stream 2 = ṁ2 × h2 = 14 × 50 = 700 kJ/(min da)

After the two streams mix into the one the resulting mass flow rate ṁ_mix = 36 + 14 = 50 kg/min

Now the energy of the resulting mass = h_mix × ṁf

From energy conservation E_mix = E1 + E2

⇒ hf × 50 = 1296 + 700

⇒ \({h_{mix}} = \frac{{1996}}{50} = 39.92 \;kJ/kg \;da\)

Concept:

Available energy or exergy is the maximum useful work that can be obtained in a process in which the system attains dead state, i.e. equilibrium with the surrounding.

The availability function of the solid substance is given as 

ϕ = U - T_oS

The exergy or available energy of the solid substance at a point is given by

ϕ₁ - ϕ_o = (U₁ - ToS₁) - (U_o - ToS_o)

∴ ϕ1 - ϕo = (U1 - Uo) + T_o(S_o - S1)

Where subscript "1" denotes the condition of the substance at any temperature and "o" denotes the condition of the substance at the atmospheric condition. 

U denotes the internal energy and S denotes the entropy.

For solid substance, the change in entropy is given by 

The change in internal energy is given by 

U1 - Uo = mC(T₁ - T_o)

Where m is the mass of the substance, C is the specific heat of the substance. 

Calculation:

Given, m = 20 kg, C = 2.8 kJ/kg°C and T₁ = 100°C = 373 K, and T_o = 20°C = 293 K

\({S_o} - {S_1} = m{C}\ln \left( {\frac{{{T_o}}}{{{T_1}}}} \right)=20×{2.8}\ln \left( {\frac{{{293}}}{{{373}}}} \right)=-13.518\;kJ\)

U1 - Uo = mC(T1 - To) = 20 × 2.8 × (373 – 293) = 4480 kJ  

∴ Available energy will be, 

ϕ1 - ϕo = (U1 - Uo) + To(So - S1)

⇒ ϕ1 - ϕo = 4480 + 293 × (- 13.518) = 519.226 kJ

Concept:

The area under the curve on T-S diagram shows the heat interaction for a reversible process

The area enclosed by the cycle on the T-S diagram gives the work interaction for the cycle

In the cyclic process

W_net = Q_net = Q_In – Q_Out

Q_out = Heat rejection at the lowest temperature in the cycle

Q_out = T_min∆S

Where ∆S is the entropy change

Calculation:

Net work is equal to the area enclosed by the cycle, i.e. area of the triangle

\({W_{net}} = \frac{1}{2} \times 4 \times 400 = 800\;kJ\)

∴ Qnet = 800 kJ

Q_out = 300 × (5-1) = 1200 kJ

So, Q_In = 800 + 1200 = 2000 kJ

The efficiency of the cycle \({\eta _{cycle}} = \frac{{{W_{net}}}}{{{Q_{net}}}} = \frac{{800}}{{2000}} = 0.4\)

There are eight properties of a system, namely pressure (p), Volume (V), temperature (T), internal energy (u), enthalpy (h), entropy (s), Helmholtz function (f) and Gibbs function (g) and h, f and g are referred to as thermodynamic potentials.

Helmholtz function F = U - TS

Gibbs free function G = H - TS

Concept:

When the water changes its phase then its pressure is the saturation pressure at that temperature.

The specific volume of a substance is defined as the volume per unit mass. 

\(v=\frac{V}{m}\)

where V is the total volume and m is the mass. 

Calculation:

Given, the mass of water in tank = 10 kg, and the volume of the tank = 0.1 m³ at 160°C.

The specific volume inside the tank is

\(v = \frac{{1{m^3}}}{{10kg}} = 0.1\frac{{{m^3}}}{{kg}}\)

For 160°, v_f = 0.001101 m³/kg and v_g = 0.3068

∴ v_f < v < v_g

Hence, water inside the tank is in saturated liquid-vapor mixture form.

\(\therefore P = (P_{sat})_{160^\circ C} = 618\;kPa\)

Concept:

The first law of thermodynamics gives the relationship between the heat transfer work transfer and the change in energy associated with a process. It is given by

dQ = ΔU + dW

where dQ is the heat interaction, dW is the work interaction and ΔU is the change in internal energy of the system.

Sign convention for the Heat and work transfer

If heat is added to the system it is treated as + ve and heat rejected by the system is treated as –ve.

If work is added to the system it is treated as –ve and work done by the system is treated as +ve.

Calculation:

Given, dW = -20 kJ/kg (work is done on the system)

dQ = -40 kJ/kg (Heat is added to the system)

Then, from first law of thermodynamics

ΔU = dQ – dW

ΔU = -40 + 20 = -20 kJ/kg

Concept:

If the balloon containing the ideal gas is initially kept in an evacuated and insulated room. Then if the balloon ruptures and the gas fills up the entire room, the process is known as free or unrestrained expansion.

Now if apply the first law of thermodynamics between the initial and final states.

\(Q=(u_2-u_1)+W\)

In this process, no work is done on or by the fluid, since the boundary of the system does not move. No heat flows to or from the fluid since the system is well insulated.

\(u_2-u_1=0\Rightarrow u_2=u_1\)

Enthalpy is given as 

h = u + Pv

For ideal gases, as we know, internal energy and enthalpy are a function of temperature only, so if internal energy U remains constant, temperature T also remains constant which means enthalpy also remains constant.

So, during the free expansion of an ideal gas, both internal energy and enthalpy remain constant.

Internal energy is the sum of potential energy of the system and the system's kinetic energy. The change in internal energy (ΔU) of a reaction is equal to the heat gained or lost (enthalpy change) in a reaction when the reaction is run at constant pressure.

Concept:

The internal energy of the gas is the sum of the internal energies of the individual gasses in the mixture.

U = U₁ + U₂ + U₃ + ……… + U_n

(Σm) C­_ve T = m₁C_v1T + m₂C_v2T + m₃C_v3T + ……… + m_nC_vnT

Therefore

\({C_{ve}} = \frac{{{\rm{\Sigma }}\left( {m \cdot {C_v}} \right)}}{{{\rm{\Sigma }}m}}\)

Similarly

The enthalpy of the gas is the sum of the enthalpies of the individual gasses in the mixture.

H = H₁ + H₂ + H₃ + ……… + H_n

(Σm) C­_Pe T = m₁C_P1T + m₂C_P2T + m₃C_P3T + ……… + m_nC_PnT

Therefore

\({C_{pe}} = \frac{{{\rm{\Sigma }}\left( {m \cdot {C_{p\;}}} \right)}}{{{\rm{\Sigma }}m}}\)

Calculation:

Given, mass fractions as m₁ = 0.5, m₂ = 0.3, m₂ = 0.2

\({C_{ve}} = \frac{{{\rm{\Sigma }}\left( {m \cdot {C_v}} \right)}}{{{\rm{\Sigma }}m}} = \frac{{0.5 \times 0.9 + 0.3 \times 1 + 0.2 \times 1.1}}{1} = 0.97\;kJ/kg.K\)

\({C_{pe}} = \frac{{{\rm{\Sigma }}\left( {m \cdot {C_{p\;}}} \right)}}{{{\rm{\Sigma }}m}} = \frac{{0.5 \times 1.2 + 0.3 \times 1.5 + 0.2 \times 1.3}}{1} = 1.31\;kJ/kg.K\)

Therefore the ratio of specific heats of the mixture

\(\frac{{{C_{pe}}}}{{{C_{ve}}}} = \frac{{1.31}}{{0.97}} = 1.35\)

electric work = I²R = 2.3 kW

On the system ⇒ W = -2.3 kW

& ΔU = -W = 2.3 kW

Concept:

The Clausius-Clapeyron equation is given by

\({\left( {\frac{{dP}}{{dT}}} \right)_{sat}} = \frac{{ds}}{{dv}} = \frac{{{s_g} - {s_f}}}{{{v_g} - {v_f}}}\)

At saturation pressure and temperature

\({s_g} - {s_f} = {s_{fg}} = \frac{{{h_{fg}}}}{T}\)

\({\left( {\frac{{dP}}{{dT}}} \right)_{sat}} = \frac{{ds}}{{dv}} = \frac{{{h_{fg}}}}{{{T_{sat}}\left( {{v_g} - {v_f}} \right)}}\)

Calculation:

 From the table

v_f = 0.001156 m³/kg, v_g = 0.12736 m³/kg

h_f = 852.43 kJ/kg, h_g = 2793.2 kJ/kg

Therefore

Fig-1 & 2 both are power cycle, so equal areas but fig-3 & 4 are reverse power cycle, so area is not meant something.

Answer A: 57.14%